It is known that the image of quadratic function y = ax & # - 178; + BX + C passes through point a (1,0) B (2, - 3) C (0,5) 1. Find the analytic expression of this quadratic function 2. The vertex coordinates of the quadratic function are obtained by the collocation method

It is known that the image of quadratic function y = ax & # - 178; + BX + C passes through point a (1,0) B (2, - 3) C (0,5) 1. Find the analytic expression of this quadratic function 2. The vertex coordinates of the quadratic function are obtained by the collocation method

[reference answer]
Three points are brought into the analytic expression of the function
a+b+c=0 ①
4a+2b+c=-3 ②
c=5 ③
The solution is a = 1, B = - 6, C = 5
So the analytic expression of the function is y = x ^ 2 - 6x + 5
y=x^2 -6x+5
=(x^2 -6x+9)-9+5
=(x-3)^2 -4
The vertex coordinates are (3, - 4)

It is proved that the function f (x) = 3x ^ 2 + 6 is an increasing function on (0, positive infinity)

x1>x2>0
f(x1)-f(x2)
=3x1^2+6-3x2^2-6
=3(x1+x2)(x1-x2)
x1>x2>0
So X1 + x2 > 0, x1-x2 > 0
So when X1 > x2 > 0, f (x1) > F (x2)
So it's an increasing function